# Pullback and Pushforward Isomorphism of Sheaves

Suppose we have two schemes $X, Y$ and a map $f\colon X\to Y$. Then we know that $\operatorname{Hom}_X(f^*\mathcal{G}, \mathcal{F})\simeq \operatorname{Hom}_Y(\mathcal{G}, f_*\mathcal{F})$, where $\mathcal{F}$ is an $\mathcal{O}_X$-module and $\mathcal{G}$ an $\mathcal{O}_Y$-module (and the Homs are in the category of $\mathcal{O}_X$-modules etc). This gives a natural map $f^* f_* \mathcal{F}\to \mathcal{F}$, just by setting $\mathcal{G}=f_* \mathcal{F}$ and looking at where the identity map goes.

Are there any well-known conditions on the map or sheaves that give this is an isomorphism? For instance, I was looking through a book and saw that the map is surjective if $\mathcal{F}$ is a very ample invertible sheaf (and maybe some more hypothesis on the map and $X$ and $Y$ were assumed as well).

This question may have general interest to students: What can be said about the canonical map

$$ρ:π∗π∗E→E?\rho: \pi^*\pi_*\mathcal{E} \rightarrow \mathcal{E}?$$

where $$π:X→S\pi: X \rightarrow S$$ any map of schemes and $$E\mathcal{E}$$ a quasi coherent $$OX\mathcal{O}_X$$-module?

Example. Let $$S:=Spec(k)S:=Spec(k)$$ with $$kk$$ a field and let $$XX$$ be a scheme of finite type over $$SS$$.
If $$E\mathcal{E}$$ is a quasi coherent $$OX\mathcal{O}_X$$-module and $$π:X→S\pi: X \rightarrow S$$ is the canonical map, it follows $$π∗π∗E→E\pi^*\pi_*\mathcal{E}\rightarrow \mathcal{E}$$
is the following map:

M1. $$ρ:OX⊗H0(X,E)→E.\rho: \mathcal{O}_X\otimes \operatorname{H}^0(X, \mathcal{E}) \rightarrow \mathcal{E}.$$

defined by $$ρ(U)(e⊗sU):=esU∈E(U)\rho(U)(e \otimes s_U ):= es_U \in \mathcal{E}(U)$$ on the open set $$UU$$. Here $$e∈OX(U)e\in \mathcal{O}_X(U)$$ and $$sUs_U$$ is the restriction of the global section $$ss$$ to the open set $$UU$$. Since $$H0(X,E) \operatorname{H}^0(X,\mathcal{E})$$ is a $$kk$$-vector space it follows $$π∗π∗E\pi^*\pi_*\mathcal{E}$$ is a free $$OX\mathcal{O}_X$$-module of rank $$dimk(H0(X,E))dim_k(\operatorname{H}^0(X,\mathcal{E}))$$.

Example.If $$E\mathcal{E}$$ is a rank $$rr$$ locally trivial $$OX\mathcal{O}_X$$-module
we may associate a “geometric vector bundle” $$π:V(E∗)→X\pi: \mathbb{V}(\mathcal{E}^*)\rightarrow X$$ to $$E\mathcal{E}$$, and the fiber $$π−1(p)≅Arκ(p)\pi^{-1}(p)\cong \mathbb{A}^r_{\kappa(p)}$$ is affine $$rr$$-space over $$κ(p)\kappa(p)$$ – the residue field of $$pp$$. The fibration $$π\pi$$ is by Hartshorne, Ex. II.5.18 a Zariski locally trivial fibration with affine spaces as fibers.
The map $$ρ\rho$$ “evaluates” a global section $$ss$$ in the fiber $$E(p)\mathcal{E}(p)$$ at a point $$p∈Xp\in X$$. We get for every $$pp$$ an element $$ρ(p)(s)∈E(p)\rho(p)(s) \in \mathcal{E}(p)$$. Here $$Arκ(p)≅Spec(Symκ(p)(E(p)∗))\mathbb{A}^r_{\kappa(p)} \cong Spec(Sym_{\kappa(p)}(\mathcal{E}(p)^*))$$

Hence the map $$ρ\rho$$ is surjective iff $$E\mathcal{E}$$ is generated by global sections. I believe the following claim is correct:

Lemma. Let $$k=Ak=A$$ be any commutative ring with $$Γ(X,E)\Gamma(X,\mathcal{E})$$ a free $$AA$$-module. If the map $$ρ\rho$$ is an isomorphism it follows $$E\mathcal{E}$$ is a free (or trivial) $$OX\mathcal{O}_X$$-module.

Proof. Let $$S:=Spec(k)S:=Spec(k)$$ and let $$π:X→S\pi: X \rightarrow S$$ be the structure morphism. There is by definition a map

M1. $$π#:OS→π∗OX\pi^{\#}: \mathcal{O}_S \rightarrow \pi_*\mathcal{O}_X$$

and this gives a canonical map

$$ρ:k:=OS(S)→Γ(X,OX)\rho: k:=\mathcal{O}_S(S) \rightarrow \Gamma(X,\mathcal{O}_X)$$.
Hence $$Γ(X,OX)\Gamma(X,\mathcal{O}_X)$$ is a $$kk$$-algebra and left $$kk$$-module.

It follows $$ρ\rho$$ gives the abelian group $$Γ(X,E)\Gamma(X, \mathcal{E})$$ the structure of a $$kk$$-module with a basis $${ei}i∈I\{e_i\}_{i\in I}$$. Hence $$π∗E\pi_*\mathcal{E}$$ is a trivial $$OS\mathcal{O}_S$$-module on the basis $${ei}i∈I\{e_i\}_{i\in I}$$. The pull back of a locally trivial $$OS\mathcal{O}_S$$-module is a locally trivial $$OX\mathcal{O}_X$$-module, and it follows there is an isomorphism

M2. $$π∗π∗E≅π∗(⊕i∈IOSei)≅⊕i∈IOXei\pi^*\pi_*\mathcal{E} \cong \pi^*(\oplus_{i \in I}\mathcal{O}_S e_i) \cong \oplus_{i\in I}\mathcal{O}_X e_i$$

since pull back commutes with direct sums and there is an isomorphism

$$π∗(OS)≅OX⊗π−1(OS)π−1(OS)≅OX\pi^*(\mathcal{O}_S)\cong \mathcal{O}_X\otimes_{\pi^{-1}(\mathcal{O}_S)} \pi^{-1}(\mathcal{O}_S) \cong \mathcal{O}_X$$

It follows $$π∗π∗E≅⊕i∈IOXei\pi^*\pi_*\mathcal{E}\cong \oplus_{i\in I} \mathcal{O}_X e_i$$ is a free $$OX\mathcal{O}_X$$-module on the set $${ei}i∈I\{e_i\}_{i\in I}$$. Hence if $$ρ\rho$$ is an isomorphism it follows $$E\mathcal{E}$$ is a free $$OX\mathcal{O}_X$$-module. QED

A comment on “free/trivial” modules. There seems to be confusion on what is meant by “free (or trivial) $$OX\mathcal{O}_X$$-module”. I use the following definition:

Definition. Given any set $$T:={ei}i∈IT:=\{e_i\}_{i\in I}$$. We say a left $$OX\mathcal{O}_X$$-module $$E\mathcal{E}$$ is “free (or trivial) on the set $$TT$$” iff for any open set $$U⊆XU \subseteq X$$ it follows

Free1. $$E(U)≅⊕i∈IOX(U)ei.\mathcal{E}(U) \cong \oplus_{i\in I}\mathcal{O}_X(U)e_i.$$

The sum is the direct sum.

It has the following restriction maps: Given $$V⊆UV\subseteq U$$ open subsets, it follows the restriction map

Free2. $$ηU,V:E(U)→E(V)\eta_{U,V}: \mathcal{E}(U) \rightarrow \mathcal{E}(V)$$

is defined by $$ηU,V(∑i∈Isiei):=⊕i∈I(si)Vei\eta_{U,V}(\sum_{i\in I} s_ie_i):=\oplus_{i\in I}(s_i)_Ve_i$$.
Here $$si∈OX(U)s_i\in \mathcal{O}_X(U)$$ and $$(si)V∈OX(V)(s_i)_V\in \mathcal{O}_X(V)$$ is the restriction of the section $$sis_i$$ to the open set $$VV$$. One checks $$E\mathcal{E}$$ is a quasi coherent $$OX\mathcal{O}_X$$-module in general.

I believe with this definition of “free/trivial”, the proof of the “Lemma/observation” above is correct. I agree – it is not difficult, but some students are not used to “sheaves” and $$OX\mathcal{O}_X$$-modules.

PS: An $$OX\mathcal{O}_X$$-module $$E\mathcal{E}$$ is called “locally trivial of rank $$rr$$” iff there is an open cover $${Ui}i∈I\{U_i\}_{i\in I}$$ with $$EUi≅(OX)rUi\mathcal{E}_{U_i} \cong (\mathcal{O}_X)_{U_i}^r$$ for all $$ii$$. The module $$E\mathcal{E}$$ is “trivial of rank $$rr$$” iff we may choose $$I:={1}I:=\{1\}$$ and $$U1:=XU_1:=X$$. Hence when people use the notion “$$E\mathcal{E}$$ is a trivial $$OX\mathcal{O}_X$$-module” this means $$E\mathcal{E}$$ is a direct sum of copies of the structure sheaf $$OX\mathcal{O}_X$$.

Example: Let $$kk$$ be a field and let $$B:=k[x],E:=⊕i∈Ik[x]ei,S:=Spec(k)B:=k[x], E:=\oplus_{i\in I}k[x]e_i, S:=Spec(k)$$ and $$C:=Spec(B)C:=Spec(B)$$

Let $$π:S→S\pi: S \rightarrow S$$ be the canonical map. It follows

$$π∗π∗E≅B⊗k(⊕i∈IBei)≅⊕i∈IB⊗kBei≅⊕i∈Ik[x,y]ei\pi^*\pi_*\mathcal{E} \cong B\otimes_k (\oplus_{i\in I}Be_i)\cong \oplus_{i\in I} B\otimes_k B e_i \cong \oplus_{i\in I} k[x,y]e_i$$

and the canonical map

$$ϕ:B⊗kE→E\phi : B\otimes_k E \rightarrow E$$ defined by $$b⊗e:=beb\otimes e:=be$$ is not an isomorphism in this case since clearly $$k[x,y]≠k[x]k[x,y] \neq k[x]$$.

“If $$F=OX\mathcal{F}=\mathcal{O}_X$$, then the stereotypical example of when this is true is when $$X=PnYX=\mathbb{P}^n_Y$$. More generally (but still with the $$F=OX\mathcal{F}=\mathcal{O}_X$$), the condition that “f is proper and all fibers are geometrically connected” may suffice, although I am not at all confident of that.”

The fiber of the morphism $$p:PnY→Yp: \mathbb{P}^n_Y \rightarrow Y$$ over a point $$ss$$ (with residue field $$κ(s):=k\kappa(s):=k$$)
is the space $$Pnk\mathbb{P}^n_{k}$$ and projective space is “geometrically connected” in general. Let $$KK$$ be the algebraic closure of $$kk$$. There is an isomorphism $$Spec(K)×Spec(k)Pnk≅PnKSpec(K) \times_{Spec(k)}\mathbb{P}^n_{k}\cong \mathbb{P}^n_{K}$$, and $$PnK\mathbb{P}^n_{K}$$ is connected for any field $$kk$$.

If $$E\mathcal{E}$$ is a finite rank locally trivial $$OX\mathcal{O}_X$$-module
we may consider $$π:X:=P(E∗)→Y\pi:X:=\mathbb{P}(\mathcal{E}^*)\rightarrow Y$$. If $$Y:=Spec(k)Y:=Spec(k)$$ with $$kk$$ and $$F\mathcal{F}$$ is a locally trivial $$OX\mathcal{O}_X$$-module that is not free it follows by the above Lemma that the map $$ρ:Γ(X,F)⊗OX→E\rho: \Gamma(X, \mathcal{F})\otimes \mathcal{O}_X \rightarrow \mathcal{E}$$ is not an isomorphism.

The answer is: The canonical map $$ρ\rho$$ is by the above argument(s) seldom an isomorphism.